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  1. Jun 15, 2014 #1

    etf

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    Hi!
    Here is my task:
    a) Calculate and sketch Iout=f(Vc2), for 0<Vc2<1.5V. (neglect Early effect)
    b) calculate output resistance for Vc2=Vc1.
    BJT's Q1 and Q2 are in forward active mode, have identical characteristics and same temperature. MOSFET M3 is in saturation. Iin=10μA, Vt=25mV, Vc1=1.25V, VA=50V.

    vdd.jpg

    My question is, what does VDD represent? There are few examples in my book where VDD is connected to some current source like this...
     
  2. jcsd
  3. Jun 16, 2014 #2

    CWatters

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    VDD will be a +ve DC supply voltage. Your text refers to VA =50V. My guess is this is a typo and VDD is meant to be VA?
     
  4. Jun 16, 2014 #3

    etf

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    I don't think so. VA is Early voltage.
     
  5. Jun 16, 2014 #4
    Generally with questions like this VDD is the biasing voltage source that puts the amplifer into the active region. My prof always used VCC for it :)
     
  6. Jun 16, 2014 #5

    CWatters

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    I think you're right.

    The actual value of VDD isn't really important in this circuit as there is a current source in series.
     
  7. Jun 16, 2014 #6

    donpacino

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    In real life that current source will be a transistor network. It it frequently represented as a current source to simplify analysis.
     
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